Question

Two satellites $\text{A}$ and $\text{B}$, have ratio of masses $3:1$ are in circular orbits of radii $r$ and $4r$. Then ratio of total mechanical energy of $\text{A}$ and $\text{B}$ is

• A. $1:3$
• B. $3:1$
• C. $3:4$
• D. $12:1$

Answer 1

The correct option is D $12:1$

The total mechanical energy of a satellite of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$ is given by
$$E=-\frac{GMm}{2r}$$ So, $E\propto \frac{m}{r}$

Given,
Ratio of the masses of the satellites, $\frac{m_A}{m_B}=\frac{3}{1}$

Ratio of the orbits of the satellites,
$\frac{r_A}{r_B}=\frac{r}{4r}=\frac{1}{4}$

Now, the ratio of the total meachanical energy,

$\frac{E_A}{E_B}=\frac{m_A}{m_B}\times \frac{r_B}{r_A}$

$\Rightarrow \frac{E_A}{E_B}=\frac{3}{1}\times \frac{4}{1}$

$\Rightarrow \frac{E_A}{E_B}=12:1$

Hence, option (d) is the correct answer.